If a group $G$ has an abelian subgroup of index $2$, then $G$ must be solvable.

Question

If a group $G$ has an abelian subgroup of index $2$, then $G$ must be solvable.

Thoughts:

I know that if $N$ is a subgroup of a group $G$ with $[G:N]=2$, then $N$ is normal.

Context:

Prof. Derek Holt makes the claim in a comment on this answer.

Please help :)

Answer 1

In order to close the question, here is @AntoineGiard's comment:

$\{e\} \triangleleft N \triangleleft G$ is a normal series such that each factor is abelian.